# What percentage of the original kinetic energy is convertible?

## Homework Statement

Ball 1 has an inertia of 0.500 kg and ball 2 has an inertia of 0.600 kg . Ball 1 is moving away from you at 5.0 m/s, and you decide to throw ball 2 at it to make it go faster. The balls collide head-on, and the coefficient of restitution for the collision is 0.95.
Part A)How fast must ball 2 be traveling in order to double the velocity of ball 1?
ANS: 9.7 m/s
Part B) What is the initial relative velocity of the two balls?
ANS: 4.7 m/s
Part C)What is their reduced inertia?
ANS: 0.27kg
Part D )What percentage of the original kinetic energy is convertible?

Kconv/Ki(x100%)

## The Attempt at a Solution

Kconv/Ki (x100%)
1.215/34.477 (x100%)= 3.5%
( every answer except Part D is correct, i'm looking for help with part D only)

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Please show how you got the result for (d), otherwise it is impossible to tell what went wrong. I get a different value for Kconv, and that is more than a single step.

Please show how you got the result for (d), otherwise it is impossible to tell what went wrong. I get a different value for Kconv, and that is more than a single step.
Ki= (0.5)(0.500)(5)^2+(0.5)(0.600)(9.7)^2
Ki=34.477

Kconv=(0.5)(0.27)(9.7-5.0)^2
=2.98251

2.98251/34.477 (x100%)= 8.6%

Ki= (0.5)(0.500)(5)^2+(0.5)(0.600)(9.7)^2
Ki=34.477

Kconv=(0.5)(0.27)(9.7-5.0)^2
=2.98251

2.98251/34.477 (x100%)= 8.6%
thats the correct answer, i found my error hours earlier and don't need help anymore